Question

You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be:

• A. 2592
• B. 2866
• C. 2976
• D. 3140

Hint:

For air-filled solenoids, permeability \( \mu_r = 1 \). Always check the unit consistency when using formulas.

Answer 1

The Correct Option is C

To determine the number of turns in the solenoid, we use the formula for the inductance \( L \) of a solenoid:

$L=\frac{\mu N_{2}A}{l}$

Where:

• $L$ is the inductance (0.016 H)
• $\mu$ is the permeability of free space (\(4π \times 10^{-7} \, \text{Tm/A}\))
• $N$ is the number of turns
• $A$ is the cross-sectional area
• $l$ is the length of the solenoid (0.81 m)

The cross-sectional area \( A \) of the solenoid is given by the formula:

$A=\mathrm{\pi r}{r}^2$

Given the radius \( r = 0.02 \, \text{m} \), we calculate:

$A=\pi (0.02{)}^2$

Substitute the given values to get \( A \):

$=3.1416\times (0.02{)}^2=1.25664\times 10{-3}^{}m²$

Substituting into the inductance formula, we have:

$\mathrm{0.016}=\frac{\left(4\pi \mathrm{\times 10}{-7}^{}\right)N^2(1.25664\times 10{-3}^{})}{0.81}$

Simplifying and solving for \( N^2 \), we get:

$N{N}^2=\frac{0.016}{\frac{0.81}{\left(}}÷(1.25664\times 10{-3}^{})$

Finally, solving for \( N \):

$N=\sqrt{\frac{0.016}{\frac{0.81}{\left(}}÷(1.25664\times 10{-3}^{})}$

After calculations, we find that \( N \approx 2976 \) turns.